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anonymous

  • 5 years ago

Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the gasoline is 6 feet, the depth is increasing at 0.2 ft/sec. How fast is the volume of gasoline changing at that instant? Round your answer to three decimal places.

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  1. anonymous
    • 5 years ago
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    Okay, you can find the volume first since you have h and r , so : \[V = \pi r^2 h\] \[= ( 3)^2 (6) \pi\] = 169.65 ft^3

  2. anonymous
    • 5 years ago
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    sstarica, isn't this the same thing we were tackling

  3. anonymous
    • 5 years ago
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    now find V' which is :\[V' = 2\pi rr'h'\]

  4. anonymous
    • 5 years ago
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    not really @iam , somehow similar :)

  5. anonymous
    • 5 years ago
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    now we have: h' = 0.2 ft/sec we need to find r'

  6. anonymous
    • 5 years ago
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    do i just plug in

  7. anonymous
    • 5 years ago
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    yes :) but we need to find r' now

  8. anonymous
    • 5 years ago
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    \[V' = 2 \pi rr' h'\] we have h' , but we need to find r' to calculate V' = rate of volume increase

  9. anonymous
    • 5 years ago
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    how do i do that? :(

  10. anonymous
    • 5 years ago
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    hold on :)

  11. anonymous
    • 5 years ago
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    thx

  12. anonymous
    • 5 years ago
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    you need to find a relationship between r and h so r/h = r'/h'

  13. anonymous
    • 5 years ago
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    thats where im having problem for all the problems

  14. anonymous
    • 5 years ago
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    cross mulitply and you'll get: r' = rh'/r

  15. anonymous
    • 5 years ago
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    sorry! wait you'll get: r' = rh'/h :)

  16. anonymous
    • 5 years ago
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    are you with me soo far?

  17. anonymous
    • 5 years ago
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    yes :)

  18. anonymous
    • 5 years ago
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    now plug : r =3 ft h = 6ft h' = 0.2ft/sec r' = (3)(0.2)/6 = 0.1 ft/sec

  19. anonymous
    • 5 years ago
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    we have found r' now, let's go back to V'\[V' = 2 \pi rr'h'\] so, V' = 2(3)(0.1)(0.2) pi = 0.377 ft^3/sec

  20. anonymous
    • 5 years ago
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    Correct me if I'm wrong please

  21. anonymous
    • 5 years ago
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    OMG this makes so much sense now

  22. anonymous
    • 5 years ago
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    but the website marked it wrong :(

  23. anonymous
    • 5 years ago
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    why?

  24. anonymous
    • 5 years ago
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    I guess there's a mistake, but that's the idea, maybe in calculation?

  25. anonymous
    • 5 years ago
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    ok

  26. anonymous
    • 5 years ago
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    did you get the idea?

  27. anonymous
    • 5 years ago
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    one qquestion

  28. anonymous
    • 5 years ago
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    sure :)

  29. anonymous
    • 5 years ago
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    where did u get \[V \prime = 2\pi r r \prime h \prime \]

  30. anonymous
    • 5 years ago
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    you derive with respect to r and h :)

  31. anonymous
    • 5 years ago
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    you want to know how is the volume increasing when both the radius and height is changing, that's the story of the related rated problem

  32. anonymous
    • 5 years ago
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    oooooh...im sorry can u show me one more problem

  33. anonymous
    • 5 years ago
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    hmmmm, let's say we have a circle of radius = 2 m, and the radius is decreasing at 0.5 m/sec. How much will the circle's area increase ?

  34. anonymous
    • 5 years ago
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    I'll give you advise, whenever you see ( m/sec) or anything with respect to time it's something' which in this case is the r' = 0.5m/sec

  35. anonymous
    • 5 years ago
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    so what is the given?

  36. anonymous
    • 5 years ago
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    we have : - R = 2 m - R' = 0.5 m/sec and we know the formula of the Area of the circle which is: A = pi r^2, right?

  37. anonymous
    • 5 years ago
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    uhhuh

  38. anonymous
    • 5 years ago
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    Now , you want to know how much is the area increasing with respect to R, since R is increasing :) so you find the derivative of it which will be : A' = 2 pi rr'

  39. anonymous
    • 5 years ago
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    so when exactly do i start pluging in numbers?

  40. anonymous
    • 5 years ago
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    right now, after you find the derived formula, start pluggingthe numbers :)

  41. anonymous
    • 5 years ago
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    okay ...these are pretty simple compared to the ones i have on hw :'(

  42. anonymous
    • 5 years ago
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    Like this problem, they want to know how much is the area increasing , we got the formula so the answer is: A' = 2(2)(0.5) pi = 2 pi = 2.68 m^2/sec :) So the area is increasing at 2.68 m^2/sec

  43. anonymous
    • 5 years ago
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    Lol, when you understand the small idea, you'll be able to figure out the bigger idea :)

  44. anonymous
    • 5 years ago
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    did you understand my problem?

  45. anonymous
    • 5 years ago
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    yes but how would i approach towards a hemisphere problem

  46. anonymous
    • 5 years ago
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    think about it, write down the given on one side, and the RTF(required to find) on the other side, then SKETCH! To test your understanding , you've got to sketch what they give you to be fully understand what they want :)

  47. anonymous
    • 5 years ago
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    to fully*

  48. anonymous
    • 5 years ago
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    1) write the given and what you need to find 2) sketch 3) write down the general formula that you're going to use :)

  49. anonymous
    • 5 years ago
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    ^_^

  50. anonymous
    • 5 years ago
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    ok :)

  51. anonymous
    • 5 years ago
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    good luck ! You're a smart student, you can figure it out :)

  52. anonymous
    • 5 years ago
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    you got the small idea, you'll be able to solve any bigger ideas now . They are not even big, they are small ideas that got stretched ^_^, don't let the problem boss u, YOU ARE THE BOSS!

  53. anonymous
    • 5 years ago
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    lol..thx

  54. anonymous
    • 5 years ago
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    np :) Give it a try once more. ^_^

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